The Fly Between Two Cars
A one-lane road is $300$ miles long. Two cars start at opposite ends and drive toward each other at constant speeds: one at $90$ mph and the other at $45$ mph. A fly starts on the slower car and flies toward the faster car at $135$ mph. The moment it reaches the faster car, it immediately turns around and flies back to the slower car at the same speed. This continues until the two cars collide.
How many miles does the fly travel in total?
Hints
- Forget the fly's zigzag path for a moment. How long is the fly in the air?
- The cars close the gap at a combined $90 + 45 = 135$ mph. The fly also moves at $135$ mph. Use $\text{distance} = \text{speed} \times \text{time}$.
- For the geometric series approach: show that after one full round trip (fly goes to the fast car and back), the remaining gap between the cars shrinks by a factor of $1/10$.
Worked Solution
How to Think About It: Before you start summing up the fly's individual back-and-forth trips, pause and think about the big picture. The fly is in the air continuously from the moment the cars start until the moment they collide. It never stops. So the total distance the fly travels is just: speed of fly $\times$ total time. The only question is: how long until the cars meet?
Quick Estimate: The two cars close the $300$-mile gap at a combined speed of $90 + 45 = 135$ mph. So they meet after $300 / 135 = 20/9 \approx 2.22$ hours. The fly moves at $135$ mph the entire time, so it covers $135 \times (300/135) = 300$ miles. Done.
Approach: That is the "trick" solution -- it avoids tracking the fly's trajectory entirely. But a classic follow-up is: can you also solve it by summing the geometric series of individual trips? Let's do that.
Formal Solution (Geometric Series Method):
Label the slow car as $S$ (starting at mile $0$, speed $45$ mph) and the fast car as $F$ (starting at mile $300$, speed $90$ mph). The fly starts on $S$.
Trip 1 (fly $\to$ $F$): The fly starts on $S$ and flies toward $F$. The closing speed between the fly and $F$ is $135 + 90 = 225$ mph across $300$ miles. Time to meet: $t_1 = 300/225 = 4/3$ hours. The fly covers $135 \times 4/3 = 180$ miles. At that moment, the gap between $S$ and $F$ is $300 - (45 + 90) \times 4/3 = 300 - 180 = 120$ miles.
Trip 2 (fly $\to$ $S$): Now the fly heads back toward $S$. Closing speed between fly and $S$ is $135 + 45 = 180$ mph across $120$ miles. Time: $120/180 = 2/3$ hours. Fly travels $135 \times 2/3 = 90$ miles. The gap shrinks by the combined car speed times time: $(90 + 45) \times 2/3 = 90$, so the new gap is $120 - 90 = 30$ miles.
After one full round trip (out and back), the fly has traveled $180 + 90 = 270$ miles, and the gap between the cars has gone from $300$ to $30$ miles -- a factor of $1/10$.
The pattern: Each full round trip shrinks the gap by a factor of $1/10$, and the fly's distance on each round trip also scales by $1/10$. So the total distance is:
$$D = 270 \sum_{i=0}^{\infty} \left(\frac{1}{10}\right)^i = 270 \times \frac{1}{1 - 1/10} = 270 \times \frac{10}{9} = 300$$
Both methods agree.
Answer: The fly travels $\mathbf{300}$ miles. The key insight is that the fly's speed equals the cars' combined closing speed, so the total distance equals the initial gap. But even without noticing that coincidence, summing the geometric series of round trips gives the same result.
Intuition
This is the classic "von Neumann fly" problem. The elegant solution is a lesson in choosing the right level of abstraction: instead of tracking every bounce of the fly, you step back and ask what total time the fly spends in the air. Since the fly never stops moving, its total distance is just speed times total time. The hard-looking infinite series collapses to a trivial calculation once you pick the right framing. In interviews, this pattern comes up constantly -- the brute-force approach (summing individual trips) works but is slow and error-prone, while the clever reframing gives the answer in one line.
The deeper lesson for quant work: before grinding through a complicated calculation, always ask whether there is a conservation law or an invariant that makes the answer obvious. Here the "invariant" is that the fly is airborne for exactly the same duration as the cars are moving. Many pricing and expected-value problems have similar shortcuts -- the hard part is recognizing when to use them.